Math Challenge - May 2021

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  • #26
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Problem 1.

We have,
$$

I=\int_0^{\infty}\int_0^{\infty}e^{-(x+y+\frac{\lambda^3}{xy})}x^{-\frac{2}{3}}y^{-\frac{1}{3}}dxdy
$$
Make the following expansion under the integral sign,
$$

e^{-\frac{\lambda^3}{xy}}=\sum_{n=0}^{\infty}(\frac{\lambda^3}{xy})^n\frac{(-1)^n}{n!}
$$
to get
$$
I=\int_0^{\infty}e^{-y}y^{-\frac{1}{3}}\int_0^{\infty}e^{-x}x^{-\frac{2}{3}}(\sum_{n=0}^{\infty}(\frac{\lambda^3}{xy})^n\frac{(-1)^n}{n!})dxdy
$$
$$
=\int_0^{\infty}e^{-y}y^{-\frac{1}{3}}\sum_{n=0}^{\infty}(\int_0^{\infty}(-1)^n(\frac{\lambda^3}{y})^n\frac{x^{-n-\frac{2}{3}}}{n!}e^{-x}dx)dy\\
$$
Each x integral term in the expansion results in a Gamma function and taking the sum out side the remaining integral we get,
$$
I=\sum_{n=0}^{\infty}(-1)^n\lambda^{3n}\frac{\Gamma (\frac{1}{3}-n)}{n!}\int_0^{\infty}e^{-y}y^{-n-\frac{1}{3}}dy
$$
Again each term y integral term in the expansion results in a Gamma function.
$$
I=\sum_{n=0}^{\infty}(-1)^n\lambda^{3n}\Gamma (\frac{2}{3}-n)\frac{\Gamma (\frac{1}{3}-n)}{n!}
$$
By iterating each term in the sum using the identity,
$$
\Gamma (n-1)=\frac{\Gamma (n)}{n-1}
$$
we find,
$$

I=1+\sum_{n=1}^{\infty}(-1)^n\frac{\lambda^{3n}3^{2n}\Gamma (\frac{1}{3}) \Gamma (\frac{2}{3}) }{\Pi_{k=1}^n (3k-1) \Pi_{k=1}^n (3k-2)n!}
$$
Let ##x=3\lambda##. On expanding each term in the sum we find,
$$


I=\Gamma (\frac{1}{3})\Gamma (\frac{2}{3})\sum_{n=0}^{\infty}(-1)^n\frac{x^{3n}}{(3n)!}
$$
Let us seek a solution for ##y(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{3n}}{(3n)!}## in the form of a second order differential equation. On expanding each term we have,
$$
y(x)= 1-\frac{x^3}{3!}+\frac{x^6}{6!}-\frac{x^9}{9!}+...
$$
$$
y'(x)=-\frac{x^2}{2!}+\frac{x^5}{5!}-\frac{x^8}{8!}+...
$$
$$
y''(x)=-x+\frac{x^4}{4!}-\frac{x^7}{7!}+\frac{x^{10}}{10!}-\frac{x^{13}}{13!}+...
$$
We observe,
$$
y''-y'+y=e^{-x}
$$
The characteristic polynomial for the homogeneous solution is,
$$
\lambda^2-\lambda + 1=0\\
$$
and thus,
$$
\lambda=\frac{1}{2}\pm \frac{i\sqrt{3}}{2}\\
$$
with the homogeneous solution,
$$
y_h(x)=e^{\frac{x}{2}}[c_1\cos(\frac{\sqrt{3}x}{2})+c_2\sin(\frac{\sqrt{3}x}{2})]
$$
The particular solution is of the form,
$$
y_p=Ce^{-x}
$$
Plugging this into the differential equation we find ##C=\frac{1}{3}##. The solution is the sum of the homogeneous part and the particular part,
$$
y(x)=y_h(x)+y_p(x)
$$
Applying the boundary conditions ##y(0)=1## and ##y'(0)=0## we find ##c_1=\frac{2}{3}## and ##c_2=0##. Thus
$$
y(x)=\frac{2}{3}e^{\frac{x}{2}}\cos(\frac{\sqrt{3}x}{2})+ \frac{1}{3}e^{-x}
$$
and we finally have,
$$
\int_0^{\infty}\int_0^{\infty}e^{-(x+y+\frac{\lambda^3}{xy})}x^{-\frac{2}{3}}y^{-\frac{1}{3}}dxdy=\Gamma (\frac{1}{3})\Gamma (\frac{2}{3})[\frac{2}{3}e^{\frac{(3\lambda)}{2}}\cos(\frac{3\sqrt{3}\lambda}{2})+ \frac{1}{3}e^{-3\lambda}]


$$
 
  • #27
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13,676
Problem 1.

We have,
$$

I=\int_0^{\infty}\int_0^{\infty}e^{-(x+y+\frac{\lambda^3}{xy})}x^{-\frac{2}{3}}y^{-\frac{1}{3}}dxdy
$$
Make the following expansion under the integral sign,
$$

e^{-\frac{\lambda^3}{xy}}=\sum_{n=0}^{\infty}(\frac{\lambda^3}{xy})^n\frac{(-1)^n}{n!}
$$
to get
$$
I=\int_0^{\infty}e^{-y}y^{-\frac{1}{3}}\int_0^{\infty}e^{-x}x^{-\frac{2}{3}}(\sum_{n=0}^{\infty}(\frac{\lambda^3}{xy})^n\frac{(-1)^n}{n!})dxdy
$$
$$
=\int_0^{\infty}e^{-y}y^{-\frac{1}{3}}\sum_{n=0}^{\infty}(\int_0^{\infty}(-1)^n(\frac{\lambda^3}{y})^n\frac{x^{-n-\frac{2}{3}}}{n!}e^{-x}dx)dy\\
$$
Each x integral term in the expansion results in a Gamma function and taking the sum out side the remaining integral we get,
$$
I=\sum_{n=0}^{\infty}(-1)^n\lambda^{3n}\frac{\Gamma (\frac{1}{3}-n)}{n!}\int_0^{\infty}e^{-y}y^{-n-\frac{1}{3}}dy
$$
Again each term y integral term in the expansion results in a Gamma function.
$$
I=\sum_{n=0}^{\infty}(-1)^n\lambda^{3n}\Gamma (\frac{2}{3}-n)\frac{\Gamma (\frac{1}{3}-n)}{n!}
$$
By iterating each term in the sum using the identity,
$$
\Gamma (n-1)=\frac{\Gamma (n)}{n-1}
$$
we find,
$$

I=1+\sum_{n=1}^{\infty}(-1)^n\frac{\lambda^{3n}3^{2n}\Gamma (\frac{1}{3}) \Gamma (\frac{2}{3}) }{\Pi_{k=1}^n (3k-1) \Pi_{k=1}^n (3k-2)n!}
$$
Let ##x=3\lambda##. On expanding each term in the sum we find,
$$


I=\Gamma (\frac{1}{3})\Gamma (\frac{2}{3})\sum_{n=0}^{\infty}(-1)^n\frac{x^{3n}}{(3n)!}
$$
Let us seek a solution for ##y(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{3n}}{(3n)!}## in the form of a second order differential equation. On expanding each term we have,
$$
y(x)= 1-\frac{x^3}{3!}+\frac{x^6}{6!}-\frac{x^9}{9!}+...
$$
$$
y'(x)=-\frac{x^2}{2!}+\frac{x^5}{5!}-\frac{x^8}{8!}+...
$$
$$
y''(x)=-x+\frac{x^4}{4!}-\frac{x^7}{7!}+\frac{x^{10}}{10!}-\frac{x^{13}}{13!}+...
$$
We observe,
$$
y''-y'+y=e^{-x}
$$
The characteristic polynomial for the homogeneous solution is,
$$
\lambda^2-\lambda + 1=0\\
$$
and thus,
$$
\lambda=\frac{1}{2}\pm \frac{i\sqrt{3}}{2}\\
$$
with the homogeneous solution,
$$
y_h(x)=e^{\frac{x}{2}}[c_1\cos(\frac{\sqrt{3}x}{2})+c_2\sin(\frac{\sqrt{3}x}{2})]
$$
The particular solution is of the form,
$$
y_p=Ce^{-x}
$$
Plugging this into the differential equation we find ##C=\frac{1}{3}##. The solution is the sum of the homogeneous part and the particular part,
$$
y(x)=y_h(x)+y_p(x)
$$
Applying the boundary conditions ##y(0)=1## and ##y'(0)=0## we find ##c_1=\frac{2}{3}## and ##c_2=0##. Thus
$$
y(x)=\frac{2}{3}e^{\frac{x}{2}}\cos(\frac{\sqrt{3}x}{2})+ \frac{1}{3}e^{-x}
$$
and we finally have,
$$
\int_0^{\infty}\int_0^{\infty}e^{-(x+y+\frac{\lambda^3}{xy})}x^{-\frac{2}{3}}y^{-\frac{1}{3}}dxdy=\Gamma (\frac{1}{3})\Gamma (\frac{2}{3})[\frac{2}{3}e^{\frac{(3\lambda)}{2}}\cos(\frac{3\sqrt{3}\lambda}{2})+ \frac{1}{3}e^{-3\lambda}]


$$
Can you simplify this expression? A lot! Will say: 1 ##e##, 1 ##\pi## , 1 ##\lambda ##, 1 ##\sqrt{{}}##, 0 ##\Gamma## I think you made a mistake. My integral is far easier. The cosine term appears to be wrong.
 
Last edited:
  • #28
benorin
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Damn I was working integral transformations not series expansions. I shouldn’t have put my pencil down!
 
  • #29
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Damn I was working integral transformations not series expansions. I shouldn’t have put my pencil down!
That is my approach, too. The series solution seems to have a mistake somewhere. There is a useful trick to solve the integral quite easily.
 
  • #30
benorin
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I get

$$I=36\sum_{k=0}^\infty (-1)^k\tfrac{\lambda ^{3k}}{(3k-1)(3k-2)}$$

as a series solution for the integral using integral transformations and series expansions and Beta functions to finish it off.
 
  • #31
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I get

$$I=36\sum_{k=0}^\infty (-1)^k\tfrac{\lambda ^{3k}}{(3k-1)(3k-2)}$$

as a series solution for the integral using integral transformations and series expansions and Beta functions to finish it off.
This is wrong. I don't know the series expansion of the solution, since it involves ##\pi## and there are so many series for ##\pi##. But I checked with ##\lambda =0## and have definitely another number.

Are we allowed to swap summation and integration if we use the series expansion? It's better to swap integration with something else.
 
  • #32
benorin
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This is wrong. I don't know the series expansion of the solution, since it involves ##\pi## and there are so many series for ##\pi##. But I checked with ##\lambda =0## and have definitely another number.

Are we allowed to swap summation and integration if we use the series expansion? It's better to swap integration with something else.
I think I see my error, the Beta function integral is not valid for ##k=0## term. So it would have been


$$I=\underbrace{4\int_0^1\int_0^1 (1-u)^{-\tfrac{1}{3}}(1-v)^{-\tfrac{2}{3}}\, du\, dv}_{=18} + 36\sum_{k=1}^\infty (-1)^k\tfrac{\lambda ^{3k}}{(3k-1)(3k-2)}$$

if this is also incorrect I guess my calculus prowess is what it was back in college :p
 
  • #33
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I think I see my error, the Beta function integral is not valid for ##k=0## term. So it would have been


$$I=4\int_0^1\int_0^1 (1-u)^{-\tfrac{1}{3}}(1-v)^{-\tfrac{2}{3}}\, du\, dv + 36\sum_{k=1}^\infty (-1)^k\tfrac{\lambda ^{3k}}{(3k-1)(3k-2)}$$

if this is also incorrect I guess my calculus prowess is what it was back in college :p
I have no idea. I am looking for a cute little solution. (cp. post #27)

The solution is a little beauty!
 
  • #35
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13,676
Refresh page and look again, I edited
@fresh_42
No. The approach via series makes it harder to see the trick. Another hint which is a good one for any integral is: eliminate what disturbs the most.
 
  • #36
benorin
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To start with separate the integrals like this

$$I=\int_0^\infty\int_0^\infty e^{-\tfrac{\lambda ^3}{xy}}x^{-\tfrac{2}{3}}y^{-\tfrac{1}{3}}\, dx\, dy \cdot \underbrace{\left( \int_0^\infty e^{-x}\, dx\right) ^2}_{=4 \Gamma ^2 (2)=4}$$

both converge as they should. Then to work on the left hand integrals by transformations. Will edit with the rest later.
 
  • #37
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This happens, dear students, if you do not note the dependencies! It has to be "for all ##\varepsilon ## there is an ##N(\varepsilon )##" and not "for all ##\varepsilon ## there is an ##N##". Sloppiness has its price!
 
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  • #38
benorin
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I was just taking a pee and realized my freshman mistake lmao I only do math once a month now and it shows
 
  • #39
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I was just taking a pee and realized my freshman mistake lmao I only do math once a month now and it shows
But if you find the idea, then it is really beautiful. @Fred Wright has been close, even with the series, but this isn't the nice solution. Forget series.
 
  • #40
Infrared
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I think Problem 2 can be done without any fiddling with commutators:

Consider the function ##\{w_1,\ldots,w_n\}\to\mathbb{Z}^n, w_i\mapsto e_i.## By the universal property of free groups, this uniquely extends to a homomorphism ##F_n\to\mathbb{Z}^n.## Note that ##x\in F_n## lies in the kernel of this map if and only if it satisfies the condition on the RHS of the equivalence (i.e. that the sums of the exponents of any fixed generator ##w_i## is zero).

Since ##\mathbb{Z}^n## is abelian, this homomorphism descends to the quotient ##F_n/[F_n,F_n]\to\mathbb{Z}_n.## We check that this map is injective by constructing a left inverse: since both groups are abelian and ##\mathbb{Z}^n## is free abelian, the function ##\{e_1,\ldots,e_n\}\to F_n/[F_n,F_n]## taking ##e_i## to the class of ##w_i## (uniquely) extends to a homomorphism ##\mathbb{Z}^n\to F_n/[F_n,F_n]## which is clearly a left inverse. (In fact, the map ##F_n/[F_n,F_n]\to\mathbb{Z}^n## is also clearly surjective so it is an isomorphism).

Now, by injectivity, ##\ker(F^n\to\mathbb{Z}^n)=\ker(F^n\to F^n/[F_n,F_n]\to\mathbb{Z}^n)=[F_n,F_n].## On the other hand, we already know that the kernel of ##F^n\to\mathbb{Z}^n## is the set of elements in ##F_n## whose exponents satisfy the given condition.

(Here, ##e_i## is the vector whose ##i##-th component is ##1## and all others are zero.)
 
Last edited:
  • #41
Infrared
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Problem 4 is quite straightforward, but the confusing part might be figuring out exactly what is being asked. I'll list everything I think we're supposed to show and perhaps someone else would like to finish it off. I'm sure @fresh_42 will let me know if I forgot something.

1. Show that ##\text{Sym}## is well-defined: Why can you divide by ##|G|##? Why is ##\text{Sym}(\varphi)## actually an element of ##\text{Hom}_{\mathbb{K}}(V,W)##?

2. Show that ##\text{Sym}## is ##\mathbb{K}##-linear.

3. Why is ##\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))## a subspace of ##\text{Hom}_{\mathbb{K}}(V,W)?##

4. Why is ##\text{Sym}(\varphi)## an element of the above space?

5. Why is ##\text{Sym}## a projection onto ##\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))##? That is, show that ##\text{Sym}^2=\text{Sym}## and that the image of ##\text{Sym}## is this space.


Also, problem ##5## confuses me. To me, it looks like ##f^n(x)## is a polynomial of odd degree ##>1## for all ##n## (not just even). So, ##f^n(x)-x## is also an odd degree polynomial and thus has a root, which is a fixed point of ##f^n##. Am I missing something? Maybe the base field is ##\mathbb{Q}## instead of ##\mathbb{R}##?
 
  • #42
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I think Problem 2 can be done without any fiddling with commutators:

Consider the function ##\{w_1,\ldots,w_n\}\to\mathbb{Z}^n, w_i\mapsto e_i.## By the universal property of free groups, this uniquely extends to a homomorphism ##F_n\to\mathbb{Z}^n.## Note that ##x\in F_n## lies in the kernel of this map if and only if it satisfies the condition on the RHS of the equivalence (i.e. that the sums of the exponents of any fixed generator ##w_i## is zero).

Since ##\mathbb{Z}^n## is abelian, this homomorphism descends to the quotient ##F_n/[F_n,F_n]\to\mathbb{Z}_n.## We check that this map is injective by constructing a left inverse: since both groups are abelian and ##\mathbb{Z}^n## is free abelian, the function ##\{e_1,\ldots,e_n\}\to F_n/[F_n,F_n]## taking ##e_i## to the class of ##w_i## (uniquely) extends to a homomorphism ##\mathbb{Z}^n\to F_n/[F_n,F_n]## which is clearly a left inverse. (In fact, the map ##F_n/[F_n,F_n]\to\mathbb{Z}^n## is also clearly surjective so it is an isomorphism).

Now, by injectivity, ##\ker(F^n\to\mathbb{Z}^n)=\ker(F^n\to F^n/[F_n,F_n]\to\mathbb{Z}^n)=[F_n,F_n].## On the other hand, we already know that the kernel of ##F^n\to\mathbb{Z}^n## is the set of elements in ##F_n## whose exponents satisfy the given condition.

(Here, ##e_i## is the vector whose ##i##-th component is ##1## and all others are zero.)
Does this count as an attempt at an answer?

The idea is correct, although you can capture all this within 4 lines. Consider for an arbitrary group ##G\longrightarrow G/[G,G],## set ##G=F_n##, done.

Since you had all points in your answer I count this as solved. But I want to mention that the direct path via induction along word length is only marginally longer and still shorter than the long answer above.
 
  • #43
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Problem 4 is quite straightforward, but the confusing part might be figuring out exactly what is being asked. I'll list everything I think we're supposed to show and perhaps someone else would like to finish it off. I'm sure @fresh_42 will let me know if I forgot something.

1. Show that ##\text{Sym}## is well-defined: Why can you divide by ##|G|##? Why is ##\text{Sym}(\varphi)## actually an element of ##\text{Hom}_{\mathbb{K}}(V,W)##?

2. Show that ##\text{Sym}## is ##\mathbb{K}##-linear.

3. Why is ##\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))## a subspace of ##\text{Hom}_{\mathbb{K}}(V,W)?##

4. Why is ##\text{Sym}(\varphi)## an element of the above space?

5. Why is ##\text{Sym}## a projection onto ##\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))##? That is, show that ##\text{Sym}^2=\text{Sym}## and that the image of ##\text{Sym}## is this space.
Not quite sure, whether your point 4 equals my 'homomorphism of representations' property, but I think so.
Also, problem ##5## confuses me. To me, it looks like ##f^n(x)## is a polynomial of odd degree ##>1## for all ##n## (not just even). So, ##f^n(x)-x## is also an odd degree polynomial and thus has a root, which is a fixed point of ##f^n##. Am I missing something? Maybe the base field is ##\mathbb{Q}## instead of ##\mathbb{R}##?
Can happen with self-constructed problems. I was so busy to find a good polynomial, that I missed the obvious. I will correct it and make it difficult.
 
  • #44
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As for #1.

I've never done a double integral before. Do you just integrate the first integral with the respect to x and then the resulting integral with respect to y?
 
  • #45
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As for #1.

I've never done a double integral before. Do you just integrate the first integral with the respect to x and then the resulting integral with respect to y?
The integration is usually inside out according to the order of the ##dx \,(1)\,dy\,(2)\,dz\,(3)## terms.

Hint: In this case, however, nested integrals are not necessary.
 
  • #46
Office_Shredder
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I feel really dumb reading #4, because I don't even see five separate claims to prove?
 
  • #47
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Problem #15

$$\begin{align}
x+y&=az\\
x-y&=bz\\
x^2+y^2&=cz
\end{align}$$

Squaring equations (1) & (2) and adding,
$$\begin{align}
2cz=(a^2+b^2)z^2\nonumber\\
z=0, \frac {2c} {a^2+b^2}\nonumber
\end{align}$$

Adding & subtracting equations (1) & (2) respectively,
$$\begin{align}
2x=z(a+b)\nonumber\\
2y=z(a-b)\nonumber
\end{align}$$
putting the value of z
$$(x,y,z) = (0,0,0) \space or \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$

edit: In the above answer, ##a^2+b^2\neq 0##
 
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  • #48
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Problem #15

$$\begin{align}
x+y&=az\\
x-y&=bz\\
x^2+y^2&=cz
\end{align}$$

Squaring equations (1) & (2) and adding,
$$\begin{align}
2cz=(a^2+b^2)z^2\nonumber\\
z=0, \frac {2c} {a^2+b^2}\nonumber
\end{align}$$

Adding & subtracting equations (1) & (2) respectively,
$$\begin{align}
2x=z(a+b)\nonumber\\
2y=z(a-b)\nonumber
\end{align}$$
putting the value of z
$$(x,y,z) = (0,0,0) \space or \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
Why does ##x=y=0## imply ##z=0##?
 
  • #49
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27
Why does x=y=0 imply z=0?
I used that z=0 imply x=y=0

as we have ##2cz=(a^2+b^2)z^2## from here we get one value of z as zero and substituting that we get x=y=0
 
  • #50
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I used that z=0 imply x=y=0

as we have ##2cz=(a^2+b^2)z^2## from here we get one value of z as zero and substituting that we get x=y=0
Sure, but what if ##z\neq 0##? You missed a possibility.
 

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